第四章习题、习题解答_图.pdf

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1、第三章习题1．系统接线如图1所示。计算f点发生三相短路时短路点电流及各支路电流。取基准功率S=100MVA。B250WVA45WVAX¢¢=0.4X¢¢=0.125dd10.5kV10.5kV60WVA300WVAu(%)=10.5u(%)=10ss115kV30km50km50kmf线路电抗为0.4W/km（图1）2．系统接线如图2所示。图中S系统的短路容量为1000MVA，求f点三相短路的短路电流。取基准功率S=100MVA。B115kV20km0.4W/km250WVAx¢¢=0.12dS系统10.5kV10kmf240WVA0.4W/kmu(%)=10s（图2）"3．

2、试计算图3所示系统中流过断路器A的最大可能电流I。115kVA10.5kV100km2´25MVA10kV，0.5kA2´20MVAcosj=0.8X=8%u=10.6%R0.4W/kmkX¢d¢=0.13300MVAS=100MVABX¢¢=0.5d（图3）4．系统接线如图4所示。求f点三相短路的短路电流。"a)次暂态电流初始值的有效值I；b)t=0.2秒时电流有效值（按同一变比法和个别变比法两种方法计算）。取基准功率S=100MVA。B6.3kV115kV2´15MVA6kV，0.6kAXd¢¢=0.125XR=10%50km0.4W/km无穷大系统f汽轮发电机2´7.5

3、MVAS=100MVABu=10.6%k（图4）习题解答1．解：S=100MVAB则：100I==0.5020kAB(110kV)3´115100I==5.4986kAB(10kV)3´10.5等值电路：XX12X8X9abXX34X5XXX131410X11cXX12XX1267fffX15XX1216ff（图）100X=0.4´=0.16001250100X=0.125´=0.2778245100X=0.098´=0.03723300100X=0.105´=0.1750460100X=0.4´30´=0.090752115100X=X=0.4´50´=0.15126721

4、15X=X+X=0.1600+0.0327=0.1927813X=X+X=0.2778+0.1750=0.4528924XX0.0907´0.151256X=X===0.03491011X+X+X0.0907+0.1512+0.1512567XX0.1512´0.151267X===0.058212X+X+X0.0907+0.1512+0.1512567X=X+X=0.1927+0.0349=0.227613810X=X+X=0.4528+0.0349=0.487714911X13X140.2276´0.4877X===0.155215X13+X140.2276+0.4877

5、X=X+X=0.0582+0.1552=0.2134161215"11I===4.6860X0.213416"I有名值=4.6860´0.5020=2.3524kA0.4877G、T支路电流I=4.6860´=3.19491110.4877+0.2276I有名值=3.1949´5.4986=17.5673kA10.2276G2、T2支路电流I2=4.6860´=1.49110.4877+0.2276I有名值=1.4911´5.4986=8.1984kA1节点电压："U=IX=4.6860´0.0582=0.2727c12U=U+IX=0.2727+3.1949´0.0349=

6、0.3842ac110U=U+IX=0.2727+1.4911´0.0349=0.3248bc211支路电流：U-U0.3842-0.3248abI===0.6556abX0.09075U0.3842aI===2.5412afX0.15126U0.3248bI===2.1479bfX0.15127I有名值=0.6556´0.5020=0.3291kAabI有名值=2.5412´0.5020=1.2757kAafI有名值=2.1479´0.5020=1.0782kAbf2.解：S=100MVAB则：100I==0.5020kAB(110kV)3´115等值电路：X1X2X3X4

7、XX67ffX5X5X8X5X9（图）S100KX===0.10001S1000B100X=0.4´20´=0.06052211510100X=´=0.04173100240100X=0.12´=0.04804250100X=0.4´10´=0.030252115X=X+X=0.1000+0.0605=0.1605612X=X+X=0.0417+0.0480=0.0897734X6X70.1605´0.0897X===0.05758X6+X70.1605+0.0897X=X+X=0.0302+0.057